Question

In ΔPQR, the bisector of ∠Q and ∠R intersect in point X. Line PX intersects side QR in point Y, then prove that `("PQ" + "PR")/"QR" = "PX"/ "XY"`.

Answer 1

Given: In ΔPQR, ray QX and ray RX are bisectors of ∠Q and ∠R respectively.

To prove: `("PQ" + "PR")/"QR" = "PX"/ "XY"`

Proof: In ΔPQY, ray QX is an angle bisector of ∠PQY.

∴ `"PX"/"XY" = "PQ"/"QY"`    ...(I) [Angle bisector theorem]

In ΔPRY, ray RX is an angle bisector of ΔPRY.

∴ `"PX"/"XY" = "PR"/"RY"`    ...(II) [Angle bisector theorem]

∴ `"PX"/"XY" = "PQ"/"QY" = "PR"/"RY"`    ...From (I) and (II)

∴ `"PX"/"XY" = ("PQ" + "PR")/("QY" + "RY")`    ...[Theorem on equal ratios]

∴ `"PX"/"XY" = ("PQ" + "PR")/"QR"`    ...(Q-Y-R)